well, we will have to chose metric or English so let's say we drop it from 1.22 meters
v = - g t = -9.81 t
h = 1.22 - (1/2)(9.81) t^2 = 0 when it hits ground
so
t^2 = 1.22/4.9
t = 0.50 seconds
so v = -9.81 * .5 = - 4.9 meters/second
car going 55,000 meters / 3600 seconds = 15.3 meters/second
so the car goes 7.65 meters during that half second
the final velocity of the phone is 4.9 m/s down and 15.3 m/s horizontal
magnitude = sqrt (15.3^2 + 4.9^2)
Mr. Tiet is driving her car at a speed of 55 km/h. he accidentally drops the phone out the window. he drops the phone from 4.0 feet.
a)How far did the car travel when the phone hit the ground?
b)What was the final velocity?
2 answers
Given: V = 55 km/h = 15.3 m/s. h = 4 Ft. = 1.21 m.
a. 0.5*g*T^2 = 1.21
0.5*9.8*T^2 = 1.21
T = 0.5 s. to hit gnd.
d = V*T = 15.3 * 0.5 = 7.65 m.
b. Y = Yo + g*T = 0 + 9.8*0.5 = 4.9 m/s.
V = Xo + Yi = 15.3 + 4.9i
V = sqrt(15.3^2 + 4.9^2) =
a. 0.5*g*T^2 = 1.21
0.5*9.8*T^2 = 1.21
T = 0.5 s. to hit gnd.
d = V*T = 15.3 * 0.5 = 7.65 m.
b. Y = Yo + g*T = 0 + 9.8*0.5 = 4.9 m/s.
V = Xo + Yi = 15.3 + 4.9i
V = sqrt(15.3^2 + 4.9^2) =